(dy)/(dx)+2xy=4x — step-by-step solution

1) Solve the linear equation

y' + 2 x y = 4 x

, where

y

is a function of

x

. The polynomial coefficients are continuous on ℝ

The form

y' + p y = q

gives

p = 2 x

and

q = 4 x

⇥1.1) Let's compute the integral:

\int(2 x)dx

⇥

2 \cdot \frac{x^{2}}{2}

⇥ℹ

2 \cdot \frac{x^{2}}{2}

\frac{2 x^{2}}{2}

‖

\frac{2 x^{2}}{2}

x^{2}

⇥

x^{2}

The integrating factor exp(∫p) is

\exp (x^{2})

, strictly positive on ℝ

Multiplying by this factor, the product rule gives

{(\exp (x^{2}) y)}^{'}

4 x \exp (x^{2})

⇥1.2) Let's compute the integral:

\int(4 x \exp (x^{2}))dx

⇥We attempt to reverse an integration: starting from

4 x \exp (x^{2})

and

\frac{d}{d x} (\exp (a)) = \exp (a)

, we differentiate using exp

⇥1.3) Let's differentiate:

\frac{d}{d x} (\exp (x^{2}))

⇥We differentiate using:

\frac{d}{d x} (\exp (f)) = \frac{d}{d x} (f) \cdot \exp (f)

(where f =

x^{2}

)

⇥

\frac{d}{d x} (x^{2}) \cdot \exp (x^{2})

⇥

2 x \cdot \exp (x^{2})

⇥You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

⇥

\frac{4}{2} \cdot \exp (x^{2})

⇥

2 \exp (x^{2})

Integrating gives

\exp (x^{2}) y = 2 \exp (x^{2}) + C

. Divide by the factor, which never vanishes;

C

is an arbitrary real constant, including zero

▶Solution:

y = \frac{2 \exp (x^{2}) + C}{\exp (x^{2})}